Evaluating a differential 2-form on test vector fields This is hopefully a quick question stemming from misunderstanding of notation. I am aiming to understand the proof of Proposition 12.17 in Lee's "Introduction to smooth manifolds" which shows that for $\omega \in \Omega^1(M) $ a 1-form on a manifold $M$ and $X, Y \in \text{Vect}(M)$ vector fields, we have $$d \omega(X, Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X, Y]).$$
We assume for a start that $\omega = u\ dv$ for smooth functions $u, v$, and calculate that $d \omega(X, Y) = (du \wedge dv)(X, Y)$. I would expect this to equal $(du)(X) (dv)(Y)$, however in Lee it is $$(du)(X) (dv)(Y) - (dv)(X) (du)(Y).$$
Where does this difference come from upon evaluation?
 A: In a basis we have $\omega = f_k \, \mathrm{d}x^k$ and $\mathrm{d}\omega = \partial_l f_k \, \mathrm{d}x^l \wedge \mathrm{d}x^k$. 
We also have
$$\begin{align}
\omega(X) & = f_k \, \mathrm{d}x^k(X) \\
& = f_k \, \mathrm{d}x^k(X^l \partial_l) \\
& = f_k X^l \, \mathrm{d}x^k(\partial_l) \\
& = f_k X^l \delta^k_l \\
& = f_k X^k
\end{align}$$
and
$$\begin{align}
(XY)(\phi) & = X(Y(\phi)) \\
& = X^k \partial_k(Y^l \partial_l \phi) \\
& = X^k (\partial_k Y^l) \partial_l \phi + X^k Y^l \partial_k \partial_l  \\
& = ((X^k \partial_k) Y^l) (\partial_l \phi) + X^k Y^l \partial_k \partial_l \phi \\
& = X(Y^l) (\partial_l \phi) + X^k Y^l \partial_k \partial_l \phi
\end{align}$$
so
$$\begin{align}
[X, Y](\phi) 
& = X(Y(\phi)) - Y(X(\phi)) \\
& = \left( X(Y^l) (\partial_l \phi) + X^k Y^l \partial_k \partial_l \phi \right) 
- \left( Y(X^l) (\partial_l \phi) + Y^k X^l \partial_k \partial_l \phi \right) \\
& = \left( X(Y^l) (\partial_l \phi) - Y(X^l) (\partial_l \phi) \right)
+ \left( X^k Y^l \partial_k \partial_l \phi 
- Y^k X^l \partial_k \partial_l \phi \right) \\
& = \left( X(Y^k) - Y(X^k) \right)  \partial_k \phi
\end{align}$$
since
$$\begin{align}
X^k Y^l \partial_k \partial_l \phi 
- Y^k X^l \partial_k \partial_l \phi
& = X^k Y^l \partial_k \partial_l \phi 
- Y^l X^l \partial_l \partial_k \phi \\
& = X^k Y^l \partial_k \partial_l \phi 
- X^l Y^l \partial_k  \partial_l\phi \\
& = 0
\end{align}$$
Thus
$$[X,Y]^k = X(Y^k) - Y(X^k).$$
This gives
$$\begin{align}
\mathrm{d}\omega(X,Y) & = \partial_l f_k (\mathrm{d}x^l \wedge \mathrm{d}x^k)(X,Y) \\
& = \partial_l f_k \, (\mathrm{d}x^l(X) \, \mathrm{d}x^k(Y) - \mathrm{d}x^k(X) \, \mathrm{d}x^l(Y)) \\
& = \partial_l f_k (X^l Y^k - X^k Y^l) \\
& = \partial_l f_k X^l Y^k - \partial_l f_k X^k Y^l \\
& = X^l (\partial_l f_k) Y^k -  Y^l (\partial_l f_k) X^k \\
& = \left( X^l \partial_l (f_k Y^k) - f_k X^l \partial_l Y^k \right) 
- \left( Y^l \partial_l (f_k X^k) - f_k Y^l \partial_l X^k  \right) \\
& = \left( X(f_k Y^k) - f_k X(Y^k) \right)
- \left( Y(f_k X^k) - f_k Y(X^k) \right) \\
& = \left( X(\omega(Y)) - f_k X(Y^k) \right)
- \left( Y(\omega(X)) - f_k Y(X^k) \right) \\
& = \left( X(\omega(Y)) - Y(\omega(X)) \right)
 -  f_k \left( X(Y^k) - Y(X^k) \right) \\
& = \left( X(\omega(Y)) - Y(\omega(X)) \right)
 -  f_k [X, Y]^k \\
& = X(\omega(Y)) - Y(\omega(X)) - \omega([X, Y])
\end{align}$$
A: You can write the wedge product as the antisymmetrization of the tensor product $du\otimes dv$, i.e. $du\wedge dv = du\otimes dv - dv \otimes du$. This then leads to the answer by the standard relation between $V\otimes V$ and $V^*\times V^*$ where $V^*$ is the dual of $V$.
